[21385]

Calculate the ratio of the cumulative stability constants for the formation of [Fe(CN)6](3-) and [Fe(CN)6](4-) ions.

Fe3+ + 1e- => Fe2+ E3=0.772 V
[Fe(CN)6]3- + 1e- => [Fe(CN)6]2- E4=0.356 V

Here is the solution:


What I don't understand is "E4=E3 + 0.059log(formation(Fe(II))/formation(Fe(III))). I understand the Nernst equation but I don't understand how it can just equate E4 with E3; they're different reactions.

Can somebody please help me? This is urgent.

Thanks a million

[Borek]

Not checked, but general guidelines that I will follow doing it will be:

Write Nernst equation for the first reaction, then solve both formation formulas for Fe(X) concentration and put these concentrations into the first Nernst equation. Now compare it with Nernst equation for the second reaction. Some expressions must be identical.

Could be I am wrong.

[Hunt]

... or you can use thermochemistry.

Delta G^o = - RT Ln K
Delta G^o  = - nFE^o

Remember that G is a state function.

Fe3+ + 1e- => Fe2+    delta G^o = - FE₃^o

Fe3+ + 6 CN^- => Fe(CN)₆3-    delta G^o = -RT Ln B_ox

Fe2+ + 6 CN^- => Fe(CN)₆4-    delta G^o = -RT Ln B_red

The target reaction :

Fe(CN)₆3- + e- => Fe(CN)₆4-    delta G = - FE₄^o

Then it's easy to show that :

- FE₄^o = - FE^o - RT Ln B_red + RT Ln B_ox

E₄^o = E₃^o + RT / F Ln (B_red / B_ox )